(a) Given, speed of a cycle = 15 km/h
and speed of scooter = 30 km/h
∴  Ratio of the speed of a cycle to the speed of the
scooter = \(\frac{15 km/h}{30 km/h}=\frac{15}{30}=\frac{1}{2}\)or 1:2
(b) We know that, 1 km = 1000 m
∴  Ratto of 5m to 10 km = \(\frac{5m}{10 km}=\frac{5 m}{10\times 1000m}\)|
\(=\frac{5}{10\times1000}=\frac{1}{2000}\)
or 1 : 2000
(c) We know that, Rs 1 = 100 paise
∴ Ratio of 50 paise to Rs 5 = \(\frac{50 paise}{Rs 5}\)
\(\frac{50 paise}{5\times 100 paise}=\frac{50}{5\times100}\)
\(=\frac{5}{5\times10}=\frac{1}{10}\) or 1:10

(i) Here, 90 parents helped their children for \(\frac{1}{2}\)h to 1\(\frac{1}{2}\) h.
Given, percentage = 30%
Let x parents be surveyed.
Then, 30% of x helped for \(\frac{1}{2}\)h to 1\(\frac{1}{2}\)h = 90
∴ 30% of x = 90
⇒ \(\frac{30}{100} \times x=90\)
⇒ \(x=\frac{90\times100}{30}=300\)
Hence, the number of parents were 300.
(ii) Here, 50% parents did not help at all.
So, number of parents did not help = 50% of 300
= \(\frac{50}{100}\times100=50\times3=150\)
Hence, 150 parents did not help.
(iii)Here, 20% parents helped for more than 1\(\frac{1}{2}\)h.
So, number of parents who helped for more than 1\(\frac{1}{2}\)h
= 20% of 300=\(\frac{20}{100} \times300=20\times3\)
= 60
Hence, 60 parents helped for more than  1\(\frac{1}{2}\)h.

Here, time period (T) =  1\(\frac{1}{2}\)yr = \(\frac{3}{2}\)yr and rate = 8%
For half-yearly, time period, T =\(\frac{3}{2}\)ｘ2 = 3
and rate = \(\frac{8}{2}\)% = 4 %

Rs 380

3%

Total number of students = 25
Students good in Mathematics = 72%
Students, who are not good in Mathematics = (100 - 72)%
= 28%
Now, number of students who are not good in Mathematics
= 28% of 25 = \(\frac{28}{100}\times25=7\)
Hence, 7 students are not good in Mathematics.

Given ratio = 2 :3
∴ Percentage =\((\frac{2}{3}\times100)=(\frac{200}{3})=66\frac{2}{3}\)%

Given ratio = 3 : 4
∴ Percentage =\((\frac{3}{4}\times100)=(3\times25)=75%\)%

We know that,
Discount = Marked price - Sale price
Sale price = Marked price - Discount
∵ Marked price = Rs 650                   [given]
Discount = 10% of Rs 650 =\(\frac{10}{100}\times 650\)=65
∴ Sale price = Marked price - Discount = Rs (650 - 65) = Rs 585

We know that,
Discount = Marked price - Sale price
Sale price = Marked price - Discount
Discount = 10% of 150 =\(\frac{10}{100}\times150\)=Rs 15
∴ Sale price of a jeans marked at  Rs 150 = 150 - 15 = Rs (150 -15) = Rs 135

We have, old price of the fan = Rs 1200
Since, the price is decreased by 15%.
∴ New price of the fan = Old price - Decreased price = 1200 -1200ｘ\(\frac{15}{100}\) = 1200-180 = Rs 1020
Alternate Method
We have, old price of the fan = Rs 1200 
∴  New Price of the fan = \(\frac{Old price\times (100-x)}{100}\)  [where, x = rate of decrease in price]
=\(\frac{1200\times(100-15)}{100}=\frac{1200\times85}{100}\) = Rs 1020  

We have, price of the article = Rs 68000
∴ New price of the article = Old price + Increased price = 68000 +68000 ｘ \(\frac{20}{100}=68000+13600=Rs 81600\)
Alternate Method
New pnce of the article = \(\frac{Old price \times (100+x)}{100}\)  [where, x = percentage of increase in price of the anicle]
\(=\frac{68000\times(100+20)}{100}=\frac{68000\times20}{100}\) = Rs 81600

600

Rs 930